Statement 17.1.2(a): The reduction of aldehydes and ketones

This reaction is also mentioned in statement 16.1.1(d) as a means of making alcohols, and you might want to refer back to that.

You should read the page about reduction of aldehydes and ketones. This contains more information than you need.

Concentrate mostly on the first section headed "Background to the reactions".

• You only need the simplified equations. Ignore any equations apart from those involving hydrogen in square brackets.

• Ignore the discussion about the conditions for the NaBH₄ reactions.

  In CIE questions, they talk about the use of an "aqueous methanolic solution of NaBH₄". In other words, for CIE purposes, the reaction is done in a mixture of water and methanol.

• For LiAlH₄, you need to know that the reaction is done initially in solution in ethoxyethane followed by treatment with acid. You will find this in the first part of the "Reaction details" section. Ignore the equations in this section.

Comparing the reduction of carbonyls with the reduction of alkenes

CIE have asked questions in the past involving the use of two different reducing agents for these bonds.

NaBH₄ and LiAlH₄ are reagents which reduce C=O double bonds, but will not reduce C=C double bonds. The normal reducing agent for C=C double bonds is hydrogen in the presence of a nickel catalyst.

Why won't NaBH₄ and LiAlH₄ reduce C=C?

This isn't on the syllabus, but CIE have expected you to know that this is the case. An outstanding student might work out the fact that you can't use these reagents to reduce C=C, but most people would just have to learn it - and since it isn't on the syllabus, why should you do that? There is no simple answer to that!

If it helps you to remember, you can think of it like this.

C=C double bonds in alkenes are reactive because of the high electron density around the pi bond. They are attacked by electrophiles - things attracted to negative charges.

But the reactive bits of NaBH₄ and LiAlH₄ lie in the negative ions, BH₄^- and AlH₄^-. These would be repelled by the pi bond, and so no reaction is possible.

Will hydrogen and nickel reduce C=O double bonds?

There is a lot of confusion about this. You will find books which say that you can reduce C=O bonds with Ni / hydrogen, and you will find books which say that you can't. CIE have twice commented on this in Examiner's Reports - once they complained that students didn't know that Ni / hydrogen could reduce C=O, and once they implied that it couldn't.

The facts would appear to be that Ni / hydrogen will reduce C=O, but more slowly than it reduces C=C.

So what do you do about this? I have had an e-mail conversation with CIE about this, and they say

"Should a question arise where both C=O and C=C could be reduced, we will mark according to the syllabus but allow candidates to say that C=O is also reduced."

That is the line you should therefore take.

• NaBH₄ and LiAlH₄ will only reduce C=O.
• Ni / hydrogen will reduce both C=C and C=O.

Statement 17.1.2(b): The reaction of aldehydes and ketones with HCN and NaCN

You should read the first part of the page about simple addition to aldehydes and ketones. You don't need the reaction with sodium hydrogensulphite.

In CIE questions they tend to talk about using hydrogen cyanide in the presence of some cyanide ions (from sodium or potassium cyanide). You will understand the reason for this when you look at the mechanism in the next statement 17.1.3.

Don't worry about the reactions mentioned in "Uses of the reaction" for now, although it might be a good idea to come back and have a look at it when you have finished all the organic chemistry. It wouldn't be too difficult to make up an exam question out of this.

Go to the Section 17 Menu . . .

To return to the list of learning outcomes in Section 17

Go to the CIE Main Menu . . .

To return to the list of all the CIE sections

Go to Chemguide Main Menu . . .

This will take you to the main part of Chemguide.